6
$\begingroup$

Some time ago I posted a question here on this forum. I would like to ask some questions regarding the way the energy per unit area between metallic plates is calculated. The full calculation is on wikipedia.

At some point in the calculation on the relevant wikipedia page (see the link above), we have the equation: $$\frac{ \langle E \rangle }{ A} = - \frac{ \hbar c \pi^2 }{6a^3}\cdot\zeta(-3) . $$

In the next step, it is written rather casually that $\zeta(-3) = - \frac{1}{120} \qquad (*) $. This is true when considering the analytic continuation of the riemann zeta function or the Ramanujan Summation method.

Therefore, it is concluded, that $$\frac{ \langle E \rangle }{A} = - \frac{ \hbar c \pi^2}{720 a^3} . $$ I am wondering under which circumstances people decided to assume the $(*)$-marked equation is 'true'. I can think of a couple of scenarios:

  1. The formula for $\frac{ \langle E \rangle }{A} $ was already derived by means of another method which did not require the use of (regularised) divergent sums. Therefore, physicists could infer that $\zeta(-3)$ had to be equal to $ - \frac{1}{120} $, making the derivation of the formula by means of this method, which does use divergent series, correct.
  2. The exact formula for $\frac{ \langle E \rangle }{A} $ was not already known. Physicist did have some data points that roughly showed them how the formula should look. Therefore, they tried some different constants for $\zeta(-3)$. At some point they guessed $\zeta(-3) = - \frac{1}{120} $, which yielded a formula that coincided with the known data points. They might have already known that $\zeta(-3) = - \frac{1}{120} $ by means of zeta function regularisation, making it easier to use this equation as a "guess" to find a suitable formula for $\frac{ \langle E \rangle }{A} $ .
  3. Some other scenario.

Which scenario roughly describes how the formula for $\frac{ \langle E \rangle }{A} $ came into existence? If it was scenario 1, which other method did physicists formerly employ to derive the formula? If it was scenario 3, how did this whole process unfold?

Thanks a lot,

Max

$\endgroup$
1

2 Answers 2

1
$\begingroup$

The zeta function is defined to be the (unique) analytic continuation of $\zeta(s):=\sum_{n=1}^\infty n^{-s}$. This implies $\zeta(-3)=1/120$.

Thus your (*) is true by definition and some theory that provides the formula
$\zeta(-n)=-B_{n+1}/(n+1)$ for natural numbers $n$. http://en.wikipedia.org/wiki/Riemann_zeta_function

If the final result agrees with the (experimental or theoretical) tradition, sloppy formal arguments like that in the Wikipedia article you cited are almost universally accepted in physics - except in mathematical physics, where the standards of rigors are much closer to those of mathematics itself.

$\endgroup$
0
$\begingroup$

Your path of thoughts seems to me a little bit flawed. In the wikipedia-article (19.3.2014) before the formal reference to the $\zeta(-3)$ the formula was derived by some conceptual arguing which ended in a formula having the infinite sum-form $\sum |n|^{-3}$. But this is only the formal description of the $\zeta(3)$.

So from some physical/logical arguing about an infinite number of standing-waves someone concluded the mathematical formula with the respective infinite number of terms which - "accidentally" - shows the form which is also that of the $\zeta()$-function.

Only after that he resorted to the equivalence, given in the mathematics of divergent series/Dirichlet-series, that this should be evaluated to $ \zeta(-3)= - \frac1{120}$ .

Why -in mathematics- this value is a good/reasonable evaluation for the $\zeta(-3)$ goes back to L.Euler and very easy arguments, for instance from the conversion of $1-8+27-64...$ (which can more easily be evaluated - with smoothing of partial sums, for instance by Euler-summation) to $1+8+27+64+...$ by his multiplication-trick and the resulting general $\eta()/\zeta()$-conversion.

Whether this purely mathematical rationale is then still reasonable in the real world of physics (and the casimir-effect and the assumption of infinitely many standing-waves) is another question and must be checked later. (I guess that the energy-formula with that value inserted has indeed already been checked to express correctly the matters which are concerned there, but I don't know that)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.